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Full text of "Treatise On Analysis Vol-Ii"

34       XII   TOPOLOGY AND TOPOLOGICAL ALGEBRA

In particular, the topology induced on the multiplicative group R*
(resp. C*) of real (resp. complex) numbers ^0 by the topology of R (resp. C)
makes R* (resp. C*) into a topological group.

If G is a topological group (in multiplicative notation), the mappings
(j^ y) i_> Xy~l and (x, y) H+ x~~ly are continuous mappings of G x G into G
(3.11.5). For each ae G, the left and right translations x\-*ax and jci- xa
are homeomorphisms of G onto G, because they are bijective and continuous
((3.20.14) and (12.5)) and so are the inverse mappings xt-*aTlx and x\r+xa~l.
For any a, b in G, the mapping xt-+axb (and in particular the inner auto-
morphism xh+axa"1) is therefore a homeomorphism of G onto G (3.11.5).
Since the mapping x\-^x"1 is bijective and equal to its inverse, it also is a
homeomorphism of G onto G.

(12.8.2)   Let G be a topological group.

(i) For each open (resp. closed) subset A of G, and each xeG, the sets
jcA, AJC and A""1 (the set of all y~~1, where y e A) are open (resp. closed) in G.

(ii) For each open set A in G and each subset B 0/G, the sets AB (the set
of all products yz, where y e A and z e B) and BA are open in G.

The assertions in (i) follow immediately from the preceding remarks, and
(ii) follows from (i), since AB = (J Az is a union of open sets.

zeB

(12.8.2.1)    On the other hand, if A and B are closed in G, it does not neces-
sarily follow that AB is closed (cf. (12.10.5)). For example, if 9 is an irrational
number, consider the subgroups Z and OZ of R. Then the subgroup Z + 0Z is
not closed in R. To see why, observe that this subgroup is denumerable, and
therefore is not the whole of R (2.2.17). Hence it is enough to prove the
following result:

(12.8.2.2)    The only closed subgroups ofR are R itself and the subgroups of
the form aZ, where a e R.

Granted this, we cannot have both 1 = na and 9 = ma with m and n
integers, because 9 is irrational, and the assertion of (12.8.2.1) then follows.

To prove (12.8.2.2), we shall first show that a subgroup H of R is either
discrete or dense in R. If H is not discrete, then for each s > 0 there exists
x 7* 0 in the set H n [-e, +e]. Since the integer multiples nx (n e Z) belong
to H, every interval of length greater than  in R contains one of these points
and therefore H is dense in R.rough f(a) and y, there is an endpoint u of ID belonging to the open